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Question: How do you prove \(\sin x + \cos x = \dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}}\) ?...

How do you prove sin⁡x+cos⁡x=2sin⁡2x−1sin⁡x−cos⁡x\sin x + \cos x = \dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}} ?

Explanation

Solution

To prove the given trigonometric identity we can start either with LHS or RHS and transform the expression in such a way that it resembles the expression in the other side. Here we are seeing that the expression in the LHS is already in a very simplified form as compared to the RHS. So we will start with the RHS and try to simplify it.

Formula used:
(a2−b2)=(a+b)(a−b)({a^2} - {b^2}) = (a + b)(a - b)
sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1

Complete step-by-step solution:
We have to prove that sin⁡x+cos⁡x=2sin⁡2x−1sin⁡x−cos⁡x\sin x + \cos x = \dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}}. We will here start with the RHS as we see that the LHS is already in a very simplified form and the RHS has better scope for being transformed.
The expression in the RHS is 2sin⁡2x−1sin⁡x−cos⁡x\dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}}. We can write it as,
2sin⁡2x−1sin⁡x−cos⁡x=sin⁡2x+sin⁡2x−1sin⁡x−cos⁡x\dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}} = \dfrac{{{{\sin }^2}x + {{\sin }^2}x - 1}}{{\sin x - \cos x}}
We know the basic trigonometric identity sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1.
⇒sin⁡2x+cos⁡2x=1 ⇒sin⁡2x−1=−cos⁡2x  \Rightarrow {\sin ^2}x + {\cos ^2}x = 1 \\\ \Rightarrow {\sin ^2}x - 1 = - {\cos ^2}x \\\
Using this in the above expression we get,
⇒2sin⁡2x−1sin⁡x−cos⁡x ⇒sin⁡2x+sin⁡2x−1sin⁡x−cos⁡x ⇒sin⁡2x−cos⁡2xsin⁡x−cos⁡x  \Rightarrow \dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}} \\\ \Rightarrow \dfrac{{{{\sin }^2}x + {{\sin }^2}x - 1}}{{\sin x - \cos x}} \\\ \Rightarrow \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x - \cos x}} \\\
In the numerator we got (sin⁡2x−cos⁡2x)({\sin ^2}x - {\cos ^2}x).
Here we use the algebraic identity, (a2−b2)=(a+b)(a−b)({a^2} - {b^2}) = (a + b)(a - b).
Thus, (sin⁡2x−cos⁡2x)=(sin⁡x+cos⁡x)(sin⁡x−cos⁡x)({\sin ^2}x - {\cos ^2}x) = (\sin x + \cos x)(\sin x - \cos x)
Using this in the above expression we get,
⇒sin⁡2x−cos⁡2xsin⁡x−cos⁡x=(sin⁡x+cos⁡x)(sin⁡x−cos⁡x)sin⁡x−cos⁡x\Rightarrow \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x - \cos x}} = \dfrac{{(\sin x + \cos x)(\sin x - \cos x)}}{{\sin x - \cos x}}
We get the expression (sin⁡x−cos⁡x)(\sin x - \cos x) both in the numerator and the denominator which gets evaluated to 11 .
Thus,
⇒(sin⁡x+cos⁡x)(sin⁡x−cos⁡x)sin⁡x−cos⁡x ⇒(sin⁡x+cos⁡x)×(sin⁡x−cos⁡x)(sin⁡x−cos⁡x) ⇒(sin⁡x+cos⁡x)×1 ⇒sin⁡x+cos⁡x  \Rightarrow \dfrac{{(\sin x + \cos x)(\sin x - \cos x)}}{{\sin x - \cos x}} \\\ \Rightarrow (\sin x + \cos x) \times \dfrac{{(\sin x - \cos x)}}{{(\sin x - \cos x)}} \\\ \Rightarrow (\sin x + \cos x) \times 1 \\\ \Rightarrow \sin x + \cos x \\\
Thus, we get RHS = sin⁡x+cos⁡x\sin x + \cos x.
Also in the trigonometric identity given in the question, we have LHS = sin⁡x+cos⁡x\sin x + \cos x.
Thus, RHS = sin⁡x+cos⁡x\sin x + \cos x = LHS.

Hence, we proved the trigonometric identity sin⁡x+cos⁡x=2sin⁡2x−1sin⁡x−cos⁡x\sin x + \cos x = \dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}}

Note: Here in the solution we will use the algebraic identities whenever it is required as algebraic identities make the calculation more simpler. Here we have used this identity (a2−b2)=(a+b)(a−b)({a^2} - {b^2}) = (a + b)(a - b) to make the calculation simple. We started with RHS for ease of our solution. We could also have started with the LHS and transformed it to resemble it to the expression in the LHS. The solution for such problems may differ from person to person as we can use different approaches. It is always a better approach to start with the more complex expression and simplify it in simpler terms.